and many more benefits!

Find us on Facebook

GMAT Club Timer Informer

Hi GMATClubber!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Show Tags

First, I will say: Questions #1 & #2, I believe, are right out as far as the GMAT -- I think they are much harder than anything the GMAT will expect you to do.

Question #3 is a perfectly legitimate GMAT question, a bit on the harder side, but well within expectations.

Question #4 could be on the GMAT -- it's at the harder end of what might be possible. ====================================================================================

Question #1:A number being successively divided by 3,5 and 8 leaves remainder 1,4 and 7 respectively. Find respective remainders if the orders of divisors is reversed?

Well, I'm going to go in reverse order. Let's say q is the final quotient. On the final division, we divided by 8 and got a remainder of 7, so the number we had before that was: 8q + 7

That, in turn, was the quotient of the previous division, where we divided by 5 with a remainder of 4. Before we divided, we must have had: 5(8q + 7) + 4.

That, in turn, was the quotient of the previous division, where we divided by 3 with a remainder of 1. Before we divided, we must have had: 3[5(8q + 7) + 4] + 1 = N, the original dividend.

I purposely left that unmultiplied out, so you could see each step's contribution. Now, multiplying out, we have N = 240q + 118.

Now, divide in the reverse order --- obviously, 3, 5, and 8 all go into 240q, so the real question is what happens when we divide the aggregate remainder term, viz, 118.

118/8 = 14, remainder = 6

14/5 = 2, remainder = 4

2/3 = 0, remainder = 2

I believe {6, 4, 2} is the remainder chain when you divide with the chain of divisors in the opposite order.

Question #2:2. When a certain number is multiplied by 13,the product consists entirely of fives. The smallest such number is:a. 41625 b. 42135 c. 42515 d. 42735e. none of the above

Well, start with the units digit --- of course, that has to be 5, so get a five in the unit's digit of the product. 5 x 13 = 65

Well, the next step is to get a 5 in the tens digit of the product. We have a 6 there already from the previous multiplication, and we can only add, not subtract, so we'll need to add a 9 --- the way to get a 9 is to multiply 13 by a 3

35 x 13 = 455

Now, we have a 4 in the hundreds place, so if we can add a one in that place we'd have a five. The way to 1 in the hundred's digit is to multiply 13 by 700

735 x 13 = 9555

Now, we have a 9 in the thousands place, so we need to add 6 --- multiply by 2000

2735 x 13 = 35555

Now, we have a 3 in the ten thousands place, so we need to add 2 --- multiply by 40000

42735 x 13 = 555555

Mirabilis dictu! As it happens, by windfall luck, we get not only a 5 in the ten thousands place, but also one in the hundred thousands place, making it a number with every digit equal to 5.

That's why D is the answer. Again, I can't imagine anyone in their right mind would expect you to slog through that without a calculator, and whatever else we say about the folks who write the GMAT, they are generally reasonable about this sort of thing.

Question #3: 3. The least number by which 72 must be multiplied in order to produce a multiple of 112, is:a. 6 b. 12 c. 14 d. 18e. none of the above

As I indicated above, this is bonafide, legitimate GMAT-type question. A questions like this do appear on the real GMAT.

The solution involves looking at the prime factorizations

72 = 8 * 9 = 2 * 2 * 2 * 3 * 3

112 = 2 * 56 = 2 * 8 * 7 = 2 * 2 * 2 * 2 * 7

So, the GCF of 72 and 112 is 2 * 2 * 2 = 8.

72 = 8 * 9

112 = 8 * 14

So, if we multiply 72 by 14, we will have a multiple of 112. In fact, in doing so, we will have created the LCM of 72 and 112. So the answer is C.